Question

Calculate the energy of 1 mole of nitrogen gas at 27o c. Compare the energy of 1 mole of hydrogen gas at this temperature with the corresponding value of nitrogen gas (Given : Boltzmann Constant (k=1.38 x 10-23 jk-1)​

Answer 1

The energy of one mole of nitrogen gas at 300 K (27°C) is equal to 6233.80 J.

• Energy of one molecule of a diatomic gas at temperature(T)  = 5×k×T/2
• Both nitrogen and hydrogen are diatomic gases.
• One mole of any gas will contain NA(Avogadro's number) molecules of that gas.
• So energy of one mole of nitrogen gas = 5×k×NA×T/2
• Putting the values of k ,NA ,T and calculating we get energy =6233.80 J
• Since both nitrogen and hydrogen are diatomic gases , the energies of one mole each of nitrogen gas and hydrogen gas will be same.

Answer 2

The ideal gas law can be written in terms of the number of molecules of gas: PV = NkT, where P is pressure, V is volume, T is temperature, N is number of molecules, and k is the Boltzmann constant k = 1.38 × 10–23 J/K. A mole is the number of atoms in a 12-g sample of carbon-12.